Hướng dẫn:

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{22.4}{56}=0,4\left( mol \right)$

$2Fe+3C{{l}_{2}}\xrightarrow{{{t}^{0}}}2FeC{{l}_{3}}$

0,4 0,6                               (mol)

Theo phương trình ta có: ${{n}_{C{{l}_{2}}}}=\frac{3.{{n}_{Fe}}}{2}=\frac{3.0,4}{2}=0,6\left( mol \right)$

$\Rightarrow {{V}_{C{{l}_{2}}}}=22,4.{{n}_{C{{l}_{2}}}}=22,4.0,6=13,44\left( l \right)$

Chọn đáp án D